Would a radio wave beam (perhaps a cm in diameter) with the same frequency as ordinary AM or FM radio waves and also the same voltage (v/m) have the same skin depth for any metal utilised to shield against it?

  • $\begingroup$ It should not in theory, since the "shielding effect" is produced by the availability of free electrons in the metal. Different metals will have different valence electrons per atom. Now, since the number of electrons is so large (10^23 in a mol which is 10^23/A atoms per gram where A is atomic mass) then you will probably need a very intense radio wave to see the difference. $\endgroup$
    – rmhleo
    Jan 16, 2016 at 22:08
  • $\begingroup$ Such a beam can not be made in the fist place. An electromagnetic wave in the AM frequency range has a wavelength of hundreds of m. If you wanted to make anything resembling a beam of that size, you would need a wavelength about ten times shorter (i.e. 1mm), so that puts you into a frequency range of, at least, 300GHz. $\endgroup$
    – CuriousOne
    Jan 17, 2016 at 2:14
  • $\begingroup$ Thanks rmhleo. Thanks CuriousOne. @CuriousOne, if an EM wave of frequency 30GHz works omni-directionally like home internet wi-fi, & there is also a separate singular beam of 30GHz (approx. 10mm), will a metal then still have the same skin depth for both wave transmissions (ie same quantity required for shielding)? (Given all other factors are equal; such as voltage, and the dimensions of the Faraday cage (ie 6 sides of 5cm2) etc.) $\endgroup$
    – tjsb55
    Jan 19, 2016 at 12:13

1 Answer 1


The skin depth of a good conductor is given by the expression $$ d = \sqrt{\frac{2}{\mu_r \mu_0 \sigma \omega}},$$ where $\omega$ is the angular frequency of the EM wave and $\sigma$ is the conductivity.

Using this we can say that the electric field strength penetrating a conductive material decays as $E = E_0 \exp(-x/d)$ and of course the penetrating power will be proportional to $E^2$ and thus will have an e-folding length which is a factor of two smaller.

Thus if $\omega$ is fixed, then the skin depth just depends on the conductivity (and relative permeability). Different metals have different conductivities; e.g. copper is about 10 times as conductive as lead, so the skin depth in copper would be smaller by a factor $\sim 3$.

Also, don't forget that there is an important reflection effect from the surfaces of conducting materials too. For a good conductor, the modulus of the transmission coefficient (from air/vacuum into the conductor) is approximately $$ |T| = 5.3\times 10^{-3} \sqrt{\frac{\mu_r \mu_0 \omega}{\sigma}}$$ Thus a more conductive material will transmit less field at the interface between the conductor and air/vacuum. This effect is more important for attenuating the E-field at low frequencies or cases where the conductor is not thick compared with the skin depth.

  • $\begingroup$ Thanks Rob Jeffries, does this formula mean the voltage (v/m) of an EM wave is completely irrelevant to the amount of metal required (skin depth) to block it? $\endgroup$
    – tjsb55
    Jan 19, 2016 at 12:14
  • $\begingroup$ @tjsb55 No, because whatever effect you are trying to block will have some threshold field to trigger it. This is an e-folding length for the field strength. See edit. $\endgroup$
    – ProfRob
    Jan 19, 2016 at 12:29
  • $\begingroup$ Thanks again things are a lot clearer now, although can you tell me what E, E0, x & μ0 mean in the above equations? $\endgroup$
    – tjsb55
    Jan 19, 2016 at 22:57
  • $\begingroup$ @tjsb55 Electric field strength, initial electric field strength and the permeability of vacuum. $\endgroup$
    – ProfRob
    Jan 19, 2016 at 23:47
  • $\begingroup$ Thanks for those, however seem to be missing the definition for the X within (-X/d) $\endgroup$
    – tjsb55
    Jan 20, 2016 at 0:39

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